THE REGULAR HEXAHEDRON OR CUBE.

In any plane construct a square ABCD such that AB is equal to a, and in a plane at right angles construct the square ABFE;

through CD draw the plane CDHG parallel to ABFE, and through EF draw the plane EFGH parallel to ABCD.

Draw the planes BCF and ADE to meet GH in G and H respectively.

Then ABCDEFGH is a cube.

PROOF.

The two planes ADHE, BCGF are parallel (Prop. 15);

hence the figure is a parallelepiped. (Def. 13.)

Each of the plane angles at A is a right angle;

hence the figure is an orthohedron. (Def. 13 and Prop. 18.) The three edges meeting at A are equal;

therefore the figure ABCDEFGH is a cube. (Def. 13.)

1

THE REGULAR OCTAHEDRON.

Construct a square ABCD whose side is equal to a; draw AC, BD meeting in 0; through O draw a straight line EOF at right angles to the plane ABCD and make OE, OF each equal to OA.

Draw the plane faces EAB, EBC, ECD, EDA, FAB, FBC, FCD, FDA.

E

C

A

Then EABCDF is a regular octahedron.

PROOF. In the right-angled triangles AOE, AOB,

AO, OE are equal to AO, OB ;

therefore AE is equal to AB.

Similarly it can be shewn that each of the other edges is equal to one of the sides of the square ABCD.

Hence each of the faces of the figure is an equilateral triangle.

The solid angle at each corner is composed of four equal solid angles, for instance the tetrahedral angle B (AECF) is made up of four trihedral angles B(AEO), B(ECO), B(CFO), and B(FAO), each of which is contained by three plane angles one of which is two-thirds and the other two each one-half of a right angle. The solid angles of the figure ABCDEF are therefore equal.

Hence the figure is a regular octahedron.

THE REGULAR DODECAHEDRON.

Let I be equal to the diagonal of a regular pentagon, whose side is equal to a, in which case,

by Ptolemy's Theorem, we have l2=al+a2.

(III. Prop. 37 B.)

Construct a cube ABCDEFGH, whose edge is equal to l.

Let O be its centre, draw from O, OL, OM, ON at right angles to the faces ABCD, ADHE, AEFB respectively.

And take OL, OM, ON each equal to (a+1); draw PLP', QMQʻ, RNR', parallel to AB, AD, AE respectively.

Take LP, LP', MQ, MQʻ, NR, NR' each equal to a.

Let l, m, n, p, p′, I, q′, r, r' be the reflexes in O of L, M, N, P, P', Q, Q', R, R'.

(See p. 620.)

Then the twenty points A, B, C, D, E, F, G, H, P, P', Q, Qʻ, R, R', p, p, q, q', r, are the corners of a regular dodecahedron.

PROOF.

The projections of AP on AB, AD, AE are equal to 1 (1 − a), § 1, ja.

Therefore AP2= † ( 1 − a )2+412+‡ a2 = }} (l2 − al+a2)= a2.

Therefore AP is equal to a.

Similarly every edge of the figure is equal to a.

(p. 533.)

(§ 4, p. 586.)

The quadrilateral APP'B has sides AP, PP', P'B, BA equal to a, a, a, l respectively and since PP' is parallel to AB it forms part of a regular pentagon whose side is equal to a.

The triangle ARB has sides AR, RB, BA equal to a, a, l respectively and therefore is equal to the remaining part of the pentagon.

Let S be the middle point of AB.

The projections of RS and RL on AD are equal to 1a and 1 (a+1), and their projections on AE are equal to (l− a) and †l,

Therefore RSL is a straight line and the figures APP'B, ARB lie in one plane and therefore the figure PARBP' is a regular pentagon. Similarly each face of the figure is a regular pentagon; and, as each solid angle of the figure is contained by three plane angles of the same size, they are all equal; therefore the figure is a regular dodecahedron.

THE REGULAR ICOSAHEDRON.

Let I be equal to the diagonal of a regular pentagon, whose side is equal to a, in which case,

by Ptolemy's Theorem, we have l2=al+a2.

(III. Prop. 37 b.)

Construct a cube ABCDEFGH whose edge is equal to 1.

Let O be the centre of the cube. Draw perpendiculars OL, OM, ON on the three faces ABCD, ADHE, AEFB, and in these faces draw lines PLP', QMQ', RNR', parallel to AB, AD, AE respectively. Take LP, LP', MQ, MQ', NR, NR' each equal to a.

Let p, p, q, q', r, r' be the reflexes of P, P', Q, Q', R, R' in the point 0. (p. 620.) The twelve points P, P', Q, Q', R, R', p, p', q, q', r, r' are the corners of a regular icosahedron.

The projections of PQ upon AB, AD, AE are equal to (-a), a, respectively.

(p. 533.)